Question

4. Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded condition (U) an

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Answer #1

Appropriate hypotheses is follows :

Ho : \mu D = 0

Ha : \mu D > 0

U A d = U - A d-dbar (d-dbar)^2
36.4 28.5 7.9 0.65 0.4225
55 20 35 27.75 770.0625
51.5 46 5.5 -1.75 3.0625
38.7 34.5 4.2 -3.05 9.3025
43.2 36.5 6.7 -0.55 0.3025
48.8 52.5 -3.7 -10.95 119.9025
25.6 26.5 -0.9 -8.15 66.4225
49.8 46.5 3.3 -3.95 15.6025
58 985.08

sum of d = 58

n = 8

sum of (d-dbar)^2 = 985.08

S = Σ(d d)2 η –1  985.08 V 8-1

s = 11.8627

Calculate T Value :

илр T = S

T 7.25 * V8 11.8627

T = 1.7286 ( calculated T value )

=n-1=8-1= 7

to.01 3.499(table ( value from T table )

Ho is passed and accepted. The average breaking load of a fabric under unabraded condition is higer than in the abraded conditions at 0.01 significance level

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